Find the area of a triangle whose sides are 34 cm, 20 cm and 42 cm. Hence, find the length of the altitude corresponding to the shortest side.

Consider 34 cm, 20 cm and 42 cm as the sides of triangle.

a = 34 cm, b = 20 cm and c = 42 cm

We know that,

Semi perimeter (s) = $\dfrac{(a + b + c)}{2}$

Substituting the values we get,

s = $\dfrac{(34 + 20 + 42)}{2} = \dfrac{96}{2}$ = 48 cm.

Area of triangle = $\sqrt{s(s - a)(s - b)(s - c)}$

Substituting values we get,

$A = \sqrt{48(48 - 34)(48 - 20)(48 - 42)} \\[1em] = \sqrt{48 \times 14 \times 28 \times 6} \\[1em] = \sqrt{112896} \\[1em] = 336 \text{ cm}^2.$

Here the shortest side of the triangle is 20 cm. Let height = h cm be the corresponding altitude.

We know that,

Area of triangle = $\dfrac{1}{2}$ × base × height

Substituting the values we get,

⇒ 336 = $\dfrac{1}{2}$ × 20 × h

⇒ h = $\dfrac{336 \times 2}{20}$

⇒ h = $\dfrac{336}{10}$

⇒ h = 33.6 cm.

Hence, area of triangle = 336 cm² and length of the altitude corresponding to the shortest side = 33.6 cm.